Practice · Common denominators reveal fraction order · Sensim Math

Variant 1 easy answer:

\frac{16}{36}

,

\frac{17}{36}

,

\frac{18}{36}

,

\frac{19}{36}

Find all fractions with a denominator of 36 that are greater than $\dfrac{5}{12}$ and less than $\dfrac{5}{9}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 36 and whose value sits strictly between 5 over 12 and 5 over 9.

Givens

The denominator must be 36.
The value must be greater than $\frac{5}{12}$ .
The value must be less than $\frac{5}{9}$ .

Unknowns

Every numerator over 36 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 36 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /36

#9 Solve an Easier Related Problem 4.NF.A.1

12 goes into 36 exactly 3 times, so multiply top and bottom by that.

\frac{5}{12} = \frac{5 \times 3}{12 \times 3} = \frac{15}{36}

The lower bound is 15 36ths.

2Convert the upper boundary to /36

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 9 goes into 36 4 times.

\frac{5}{9} = \frac{5 \times 4}{9 \times 4} = \frac{20}{36}

The upper bound is 20 36ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 36ths, so ordering them is ordering their numerators.

\frac{15}{36} < \frac{\square}{36} < \frac{20}{36}

The numerator has to sit strictly between 15 and 20.

4List the whole-number numerators between 15 and 20

#2 Make a Systematic List 4.NF.A.2

Neither 15 nor 20 counts, since both comparisons are strict.

\frac{16}{36}, \frac{17}{36}, \frac{18}{36}, \frac{19}{36}

4 fractions fit.

Answer:

\frac{16}{36}

,

\frac{17}{36}

,

\frac{18}{36}

,

\frac{19}{36}

4 · Reviewdoes it hold up?

Check the ends: 15/36 equals the lower bound exactly, so it is out, and 20/36 equals the upper bound, also out. The 4 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 4 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 36.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 2 easy answer:

\frac{11}{40}

,

\frac{12}{40}

,

\frac{13}{40}

,

\frac{14}{40}

Find all fractions with a denominator of 40 that are greater than $\dfrac{1}{4}$ and less than $\dfrac{3}{8}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 40 and whose value sits strictly between 1 over 4 and 3 over 8.

Givens

The denominator must be 40.
The value must be greater than $\frac{1}{4}$ .
The value must be less than $\frac{3}{8}$ .

Unknowns

Every numerator over 40 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 40 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /40

#9 Solve an Easier Related Problem 4.NF.A.1

4 goes into 40 exactly 10 times, so multiply top and bottom by that.

\frac{1}{4} = \frac{1 \times 10}{4 \times 10} = \frac{10}{40}

The lower bound is 10 40ths.

2Convert the upper boundary to /40

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 8 goes into 40 5 times.

\frac{3}{8} = \frac{3 \times 5}{8 \times 5} = \frac{15}{40}

The upper bound is 15 40ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 40ths, so ordering them is ordering their numerators.

\frac{10}{40} < \frac{\square}{40} < \frac{15}{40}

The numerator has to sit strictly between 10 and 15.

4List the whole-number numerators between 10 and 15

#2 Make a Systematic List 4.NF.A.2

Neither 10 nor 15 counts, since both comparisons are strict.

\frac{11}{40}, \frac{12}{40}, \frac{13}{40}, \frac{14}{40}

4 fractions fit.

Answer:

\frac{11}{40}

,

\frac{12}{40}

,

\frac{13}{40}

,

\frac{14}{40}

4 · Reviewdoes it hold up?

Check the ends: 10/40 equals the lower bound exactly, so it is out, and 15/40 equals the upper bound, also out. The 4 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 4 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 40.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 3 easy answer:

\frac{17}{48}

,

\frac{18}{48}

,

\frac{19}{48}

Find all fractions with a denominator of 48 that are greater than $\dfrac{1}{3}$ and less than $\dfrac{5}{12}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 48 and whose value sits strictly between 1 over 3 and 5 over 12.

Givens

The denominator must be 48.
The value must be greater than $\frac{1}{3}$ .
The value must be less than $\frac{5}{12}$ .

Unknowns

Every numerator over 48 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 48 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /48

#9 Solve an Easier Related Problem 4.NF.A.1

3 goes into 48 exactly 16 times, so multiply top and bottom by that.

\frac{1}{3} = \frac{1 \times 16}{3 \times 16} = \frac{16}{48}

The lower bound is 16 48ths.

2Convert the upper boundary to /48

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 12 goes into 48 4 times.

\frac{5}{12} = \frac{5 \times 4}{12 \times 4} = \frac{20}{48}

The upper bound is 20 48ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 48ths, so ordering them is ordering their numerators.

\frac{16}{48} < \frac{\square}{48} < \frac{20}{48}

The numerator has to sit strictly between 16 and 20.

4List the whole-number numerators between 16 and 20

#2 Make a Systematic List 4.NF.A.2

Neither 16 nor 20 counts, since both comparisons are strict.

\frac{17}{48}, \frac{18}{48}, \frac{19}{48}

3 fractions fit.

Answer:

\frac{17}{48}

,

\frac{18}{48}

,

\frac{19}{48}

4 · Reviewdoes it hold up?

Check the ends: 16/48 equals the lower bound exactly, so it is out, and 20/48 equals the upper bound, also out. The 3 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 3 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 48.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 4 easy answer:

\frac{13}{54}

,

\frac{14}{54}

Find all fractions with a denominator of 54 that are greater than $\dfrac{2}{9}$ and less than $\dfrac{5}{18}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 54 and whose value sits strictly between 2 over 9 and 5 over 18.

Givens

The denominator must be 54.
The value must be greater than $\frac{2}{9}$ .
The value must be less than $\frac{5}{18}$ .

Unknowns

Every numerator over 54 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 54 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /54

#9 Solve an Easier Related Problem 4.NF.A.1

9 goes into 54 exactly 6 times, so multiply top and bottom by that.

\frac{2}{9} = \frac{2 \times 6}{9 \times 6} = \frac{12}{54}

The lower bound is 12 54ths.

2Convert the upper boundary to /54

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 18 goes into 54 3 times.

\frac{5}{18} = \frac{5 \times 3}{18 \times 3} = \frac{15}{54}

The upper bound is 15 54ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 54ths, so ordering them is ordering their numerators.

\frac{12}{54} < \frac{\square}{54} < \frac{15}{54}

The numerator has to sit strictly between 12 and 15.

4List the whole-number numerators between 12 and 15

#2 Make a Systematic List 4.NF.A.2

Neither 12 nor 15 counts, since both comparisons are strict.

\frac{13}{54}, \frac{14}{54}

2 fractions fit.

Answer:

\frac{13}{54}

,

\frac{14}{54}

4 · Reviewdoes it hold up?

Check the ends: 12/54 equals the lower bound exactly, so it is out, and 15/54 equals the upper bound, also out. The 2 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 2 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 54.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 5 medium answer:

\frac{25}{60}

,

\frac{26}{60}

,

\frac{27}{60}

,

\frac{28}{60}

,

\frac{29}{60}

Find all fractions with a denominator of 60 that are greater than $\dfrac{2}{5}$ and less than $\dfrac{1}{2}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 60 and whose value sits strictly between 2 over 5 and 1 over 2.

Givens

The denominator must be 60.
The value must be greater than $\frac{2}{5}$ .
The value must be less than $\frac{1}{2}$ .

Unknowns

Every numerator over 60 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 60 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /60

#9 Solve an Easier Related Problem 4.NF.A.1

5 goes into 60 exactly 12 times, so multiply top and bottom by that.

\frac{2}{5} = \frac{2 \times 12}{5 \times 12} = \frac{24}{60}

The lower bound is 24 60ths.

2Convert the upper boundary to /60

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 2 goes into 60 30 times.

\frac{1}{2} = \frac{1 \times 30}{2 \times 30} = \frac{30}{60}

The upper bound is 30 60ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 60ths, so ordering them is ordering their numerators.

\frac{24}{60} < \frac{\square}{60} < \frac{30}{60}

The numerator has to sit strictly between 24 and 30.

4List the whole-number numerators between 24 and 30

#2 Make a Systematic List 4.NF.A.2

Neither 24 nor 30 counts, since both comparisons are strict.

\frac{25}{60}, \frac{26}{60}, \frac{27}{60}, \frac{28}{60}, \frac{29}{60}

5 fractions fit.

Answer:

\frac{25}{60}

,

\frac{26}{60}

,

\frac{27}{60}

,

\frac{28}{60}

,

\frac{29}{60}

4 · Reviewdoes it hold up?

Check the ends: 24/60 equals the lower bound exactly, so it is out, and 30/60 equals the upper bound, also out. The 5 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 5 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 60.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 6 medium answer:

\frac{22}{63}

,

\frac{23}{63}

,

\frac{24}{63}

,

\frac{25}{63}

,

\frac{26}{63}

Find all fractions with a denominator of 63 that are greater than $\dfrac{1}{3}$ and less than $\dfrac{3}{7}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 63 and whose value sits strictly between 1 over 3 and 3 over 7.

Givens

The denominator must be 63.
The value must be greater than $\frac{1}{3}$ .
The value must be less than $\frac{3}{7}$ .

Unknowns

Every numerator over 63 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 63 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /63

#9 Solve an Easier Related Problem 4.NF.A.1

3 goes into 63 exactly 21 times, so multiply top and bottom by that.

\frac{1}{3} = \frac{1 \times 21}{3 \times 21} = \frac{21}{63}

The lower bound is 21 63ths.

2Convert the upper boundary to /63

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 7 goes into 63 9 times.

\frac{3}{7} = \frac{3 \times 9}{7 \times 9} = \frac{27}{63}

The upper bound is 27 63ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 63ths, so ordering them is ordering their numerators.

\frac{21}{63} < \frac{\square}{63} < \frac{27}{63}

The numerator has to sit strictly between 21 and 27.

4List the whole-number numerators between 21 and 27

#2 Make a Systematic List 4.NF.A.2

Neither 21 nor 27 counts, since both comparisons are strict.

\frac{22}{63}, \frac{23}{63}, \frac{24}{63}, \frac{25}{63}, \frac{26}{63}

5 fractions fit.

Answer:

\frac{22}{63}

,

\frac{23}{63}

,

\frac{24}{63}

,

\frac{25}{63}

,

\frac{26}{63}

4 · Reviewdoes it hold up?

Check the ends: 21/63 equals the lower bound exactly, so it is out, and 27/63 equals the upper bound, also out. The 5 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 5 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 63.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 7 medium answer:

\frac{28}{72}

,

\frac{29}{72}

,

\frac{30}{72}

,

\frac{31}{72}

Find all fractions with a denominator of 72 that are greater than $\dfrac{3}{8}$ and less than $\dfrac{4}{9}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 72 and whose value sits strictly between 3 over 8 and 4 over 9.

Givens

The denominator must be 72.
The value must be greater than $\frac{3}{8}$ .
The value must be less than $\frac{4}{9}$ .

Unknowns

Every numerator over 72 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 72 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /72

#9 Solve an Easier Related Problem 4.NF.A.1

8 goes into 72 exactly 9 times, so multiply top and bottom by that.

\frac{3}{8} = \frac{3 \times 9}{8 \times 9} = \frac{27}{72}

The lower bound is 27 72ths.

2Convert the upper boundary to /72

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 9 goes into 72 8 times.

\frac{4}{9} = \frac{4 \times 8}{9 \times 8} = \frac{32}{72}

The upper bound is 32 72ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 72ths, so ordering them is ordering their numerators.

\frac{27}{72} < \frac{\square}{72} < \frac{32}{72}

The numerator has to sit strictly between 27 and 32.

4List the whole-number numerators between 27 and 32

#2 Make a Systematic List 4.NF.A.2

Neither 27 nor 32 counts, since both comparisons are strict.

\frac{28}{72}, \frac{29}{72}, \frac{30}{72}, \frac{31}{72}

4 fractions fit.

Answer:

\frac{28}{72}

,

\frac{29}{72}

,

\frac{30}{72}

,

\frac{31}{72}

4 · Reviewdoes it hold up?

Check the ends: 27/72 equals the lower bound exactly, so it is out, and 32/72 equals the upper bound, also out. The 4 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 4 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 72.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 8 medium answer:

\frac{31}{80}

,

\frac{32}{80}

,

\frac{33}{80}

,

\frac{34}{80}

,

\frac{35}{80}

Find all fractions with a denominator of 80 that are greater than $\dfrac{3}{8}$ and less than $\dfrac{9}{20}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 80 and whose value sits strictly between 3 over 8 and 9 over 20.

Givens

The denominator must be 80.
The value must be greater than $\frac{3}{8}$ .
The value must be less than $\frac{9}{20}$ .

Unknowns

Every numerator over 80 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 80 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /80

#9 Solve an Easier Related Problem 4.NF.A.1

8 goes into 80 exactly 10 times, so multiply top and bottom by that.

\frac{3}{8} = \frac{3 \times 10}{8 \times 10} = \frac{30}{80}

The lower bound is 30 80ths.

2Convert the upper boundary to /80

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 20 goes into 80 4 times.

\frac{9}{20} = \frac{9 \times 4}{20 \times 4} = \frac{36}{80}

The upper bound is 36 80ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 80ths, so ordering them is ordering their numerators.

\frac{30}{80} < \frac{\square}{80} < \frac{36}{80}

The numerator has to sit strictly between 30 and 36.

4List the whole-number numerators between 30 and 36

#2 Make a Systematic List 4.NF.A.2

Neither 30 nor 36 counts, since both comparisons are strict.

\frac{31}{80}, \frac{32}{80}, \frac{33}{80}, \frac{34}{80}, \frac{35}{80}

5 fractions fit.

Answer:

\frac{31}{80}

,

\frac{32}{80}

,

\frac{33}{80}

,

\frac{34}{80}

,

\frac{35}{80}

4 · Reviewdoes it hold up?

Check the ends: 30/80 equals the lower bound exactly, so it is out, and 36/80 equals the upper bound, also out. The 5 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 5 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 80.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 9 hard answer:

\frac{43}{84}

,

\frac{44}{84}

,

\frac{45}{84}

,

\frac{46}{84}

,

\frac{47}{84}

Find all fractions with a denominator of 84 that are greater than $\dfrac{1}{2}$ and less than $\dfrac{4}{7}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 84 and whose value sits strictly between 1 over 2 and 4 over 7.

Givens

The denominator must be 84.
The value must be greater than $\frac{1}{2}$ .
The value must be less than $\frac{4}{7}$ .

Unknowns

Every numerator over 84 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 84 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /84

#9 Solve an Easier Related Problem 4.NF.A.1

2 goes into 84 exactly 42 times, so multiply top and bottom by that.

\frac{1}{2} = \frac{1 \times 42}{2 \times 42} = \frac{42}{84}

The lower bound is 42 84ths.

2Convert the upper boundary to /84

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 7 goes into 84 12 times.

\frac{4}{7} = \frac{4 \times 12}{7 \times 12} = \frac{48}{84}

The upper bound is 48 84ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 84ths, so ordering them is ordering their numerators.

\frac{42}{84} < \frac{\square}{84} < \frac{48}{84}

The numerator has to sit strictly between 42 and 48.

4List the whole-number numerators between 42 and 48

#2 Make a Systematic List 4.NF.A.2

Neither 42 nor 48 counts, since both comparisons are strict.

\frac{43}{84}, \frac{44}{84}, \frac{45}{84}, \frac{46}{84}, \frac{47}{84}

5 fractions fit.

Answer:

\frac{43}{84}

,

\frac{44}{84}

,

\frac{45}{84}

,

\frac{46}{84}

,

\frac{47}{84}

4 · Reviewdoes it hold up?

Check the ends: 42/84 equals the lower bound exactly, so it is out, and 48/84 equals the upper bound, also out. The 5 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 5 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 84.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 10 hard answer:

\frac{41}{90}

,

\frac{42}{90}

,

\frac{43}{90}

,

\frac{44}{90}

Find all fractions with a denominator of 90 that are greater than $\dfrac{4}{9}$ and less than $\dfrac{1}{2}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 90 and whose value sits strictly between 4 over 9 and 1 over 2.

Givens

The denominator must be 90.
The value must be greater than $\frac{4}{9}$ .
The value must be less than $\frac{1}{2}$ .

Unknowns

Every numerator over 90 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 90 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /90

#9 Solve an Easier Related Problem 4.NF.A.1

9 goes into 90 exactly 10 times, so multiply top and bottom by that.

\frac{4}{9} = \frac{4 \times 10}{9 \times 10} = \frac{40}{90}

The lower bound is 40 90ths.

2Convert the upper boundary to /90

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 2 goes into 90 45 times.

\frac{1}{2} = \frac{1 \times 45}{2 \times 45} = \frac{45}{90}

The upper bound is 45 90ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 90ths, so ordering them is ordering their numerators.

\frac{40}{90} < \frac{\square}{90} < \frac{45}{90}

The numerator has to sit strictly between 40 and 45.

4List the whole-number numerators between 40 and 45

#2 Make a Systematic List 4.NF.A.2

Neither 40 nor 45 counts, since both comparisons are strict.

\frac{41}{90}, \frac{42}{90}, \frac{43}{90}, \frac{44}{90}

4 fractions fit.

Answer:

\frac{41}{90}

,

\frac{42}{90}

,

\frac{43}{90}

,

\frac{44}{90}

4 · Reviewdoes it hold up?

Check the ends: 40/90 equals the lower bound exactly, so it is out, and 45/90 equals the upper bound, also out. The 4 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 4 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 90.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 11 hard answer:

\frac{51}{100}

,

\frac{52}{100}

,

\frac{53}{100}

,

\frac{54}{100}

Find all fractions with a denominator of 100 that are greater than $\dfrac{1}{2}$ and less than $\dfrac{11}{20}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 100 and whose value sits strictly between 1 over 2 and 11 over 20.

Givens

The denominator must be 100.
The value must be greater than $\frac{1}{2}$ .
The value must be less than $\frac{11}{20}$ .

Unknowns

Every numerator over 100 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 100 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /100

#9 Solve an Easier Related Problem 4.NF.A.1

2 goes into 100 exactly 50 times, so multiply top and bottom by that.

\frac{1}{2} = \frac{1 \times 50}{2 \times 50} = \frac{50}{100}

The lower bound is 50 100ths.

2Convert the upper boundary to /100

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 20 goes into 100 5 times.

\frac{11}{20} = \frac{11 \times 5}{20 \times 5} = \frac{55}{100}

The upper bound is 55 100ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 100ths, so ordering them is ordering their numerators.

\frac{50}{100} < \frac{\square}{100} < \frac{55}{100}

The numerator has to sit strictly between 50 and 55.

4List the whole-number numerators between 50 and 55

#2 Make a Systematic List 4.NF.A.2

Neither 50 nor 55 counts, since both comparisons are strict.

\frac{51}{100}, \frac{52}{100}, \frac{53}{100}, \frac{54}{100}

4 fractions fit.

Answer:

\frac{51}{100}

,

\frac{52}{100}

,

\frac{53}{100}

,

\frac{54}{100}

4 · Reviewdoes it hold up?

Check the ends: 50/100 equals the lower bound exactly, so it is out, and 55/100 equals the upper bound, also out. The 4 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 4 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 100.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Variant 12 hard answer:

\frac{51}{120}

,

\frac{52}{120}

,

\frac{53}{120}

,

\frac{54}{120}

,

\frac{55}{120}

Find all fractions with a denominator of 120 that are greater than $\dfrac{5}{12}$ and less than $\dfrac{7}{15}$ .

Show solution

1 · Understandwhat's really being asked

We want every fraction whose denominator is 120 and whose value sits strictly between 5 over 12 and 7 over 15.

Givens

The denominator must be 120.
The value must be greater than $\frac{5}{12}$ .
The value must be less than $\frac{7}{15}$ .

Unknowns

Every numerator over 120 that fits.

Constraints

Both comparisons are strict, so neither boundary counts.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #9 Solve an Easier Related Problem

Three different denominators cannot be compared directly. Put both bounds over 120 as well; then all three count pieces of the same size and the answer is just the numerators in between.

3 · Execute4 carry out the plan

1Convert the lower boundary to /120

#9 Solve an Easier Related Problem 4.NF.A.1

12 goes into 120 exactly 10 times, so multiply top and bottom by that.

\frac{5}{12} = \frac{5 \times 10}{12 \times 10} = \frac{50}{120}

The lower bound is 50 120ths.

2Convert the upper boundary to /120

#9 Solve an Easier Related Problem 4.NF.A.1

The same move on the other bound: 15 goes into 120 8 times.

\frac{7}{15} = \frac{7 \times 8}{15 \times 8} = \frac{56}{120}

The upper bound is 56 120ths.

3Compare using equal-size pieces

#2 Make a Systematic List 4.NF.A.2

All three fractions now count 120ths, so ordering them is ordering their numerators.

\frac{50}{120} < \frac{\square}{120} < \frac{56}{120}

The numerator has to sit strictly between 50 and 56.

4List the whole-number numerators between 50 and 56

#2 Make a Systematic List 4.NF.A.2

Neither 50 nor 56 counts, since both comparisons are strict.

\frac{51}{120}, \frac{52}{120}, \frac{53}{120}, \frac{54}{120}, \frac{55}{120}

5 fractions fit.

Answer:

\frac{51}{120}

,

\frac{52}{120}

,

\frac{53}{120}

,

\frac{54}{120}

,

\frac{55}{120}

4 · Reviewdoes it hold up?

Check the ends: 50/120 equals the lower bound exactly, so it is out, and 56/120 equals the upper bound, also out. The 5 between them are all strictly inside.

Another way: Cross-multiplying each candidate against both bounds gives the same list, at 5 times two multiplications rather than two conversions.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both bounds over 120.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Ordering by numerator once the denominators match.

💡Takeaway. Give every fraction the same denominator and comparing them becomes counting -- the numerators do all the ordering.

Common denominators reveal fraction order · 12 practice problems

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